Gravitational Force-Need Help Fast

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8*10^7 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 12.2 m/s^2.

What is the mass of the planet? in kg

What is the mass of the star? in kg

Equations:
Mass(of planet)= g*R^2/G

3. The attempt at a solution

Ok, so I calculated the mass of the planet to be 1.48*10^25 kg using the above equation. M=[(12.2m/s^2)*(9*10^6m)^2]/6.67*10^-11 Nm^2/kg^2.

Now I am stuck with how to find the mass of the star and which equation to use to get it..

Staff: Mentor

It states that the planet orbits 2.2*10^11 m from its star. This would them be the distance, correct??

Yes, that is a or r in Kepler's forumula relating period with distance for one mass orbiting another mass, e.g. moon around a planet or planet about a star.

In the simplest form, one may assume that the mass of the star greatly exceeds the mass of the planet.

So [tex]T^2\,=\,\frac{4{\pi^2}{a^3}}{GM}[/tex], which can be rearranged to get

[tex]M\,=\,\frac{4{\pi^2}{a^3}}{GT^2}[/tex]

So substitute in the appropriate numbers

G = 6.67 x 10-11 N m2/kg2, the universal gravitational constant,

a = 2.2 x 1011 m

T = 402 d * 24 h/d * 3600 s/h = 3.47328 x 107 s

Now compare the mass of the star with the mass of the planet. The mass calculated for the star might need adjusting for the mass of the planet since M is the sum of the masses, but if M >> m(planet), the M is approximately the mass of the star.

M=[(2pir^3/2)^2]/GT

This is not correct. Be careful about moving terms and exponents. T should be squared and if one brings r^3 inside the parentheses and squares those terms, then one must use r^(3/2) within the parentheses.